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Q18CQ

Expert-verifiedFound in: Page 662

Book edition
1st Edition

Author(s)
Paul Peter Urone

Pages
1272 pages

ISBN
9781938168000

**If the electric field lines in the figure above were perpendicular to the object, would it necessarily be a conductor? Explain.**

When electric field lines are perpendicular to the object, the object must be a conductor.

Gauss stated that the electric flux ${\mathit{\phi}}$ through a hypothetical Gaussian surface equals to $\raisebox{1ex}{$\mathbf{1}$}\!\left/ \!\raisebox{-1ex}{${\mathbf{\epsilon}}_{\mathbf{\circ}}$}\right.$ times of the total charge enclosed inside the hypothetical Gaussian surface. Mathematically,

${\mathit{\phi}}{\mathbf{=}}\frac{{\mathbf{q}}_{\mathbf{e}\mathbf{n}\mathbf{c}}}{{\mathbf{\epsilon}}_{\mathbf{\circ}}}{\mathbf{.}}{\mathbf{.}}{\mathbf{.}}{\mathbf{.}}{\mathbf{.}}{\mathbf{.}}{\mathbf{.}}{\mathbf{.}}{\mathbf{.}}{\mathbf{.}}{\mathbf{(}}{\mathbf{1}}{\mathbf{.}}{\mathbf{1}}{\mathbf{)}}$

Here, ${{\mathit{q}}}_{\mathbf{e}\mathbf{n}\mathbf{c}}$ is the charge enclosed inside the Gaussian surface, and ${{\mathit{\epsilon}}}_{{\mathbf{\circ}}}$ is the permittivity of the free space.

When some charge is given to the conductor, the charge distributes itself on the surface of the conductor in order to reduce the repulsion between them. This mutual repulsion can only be minimized if there is no component of the field that is applying force on other charges. This is only possible if the field lines are perpendicular to the surface charge distribution.

Thus. The electric field for a conductor is perpendicular to its surface.

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